This question is a particular take on the following theme. Suppose $A$ and $B$ are two notions of "large subset of $\omega^\omega$;" when is there a uniform method for turning an element of $A$ into an element of $B$?

We work in ZF (although results under strengthenings of ZF are also interesting).

For $U, V\subseteq\mathcal{P}(\omega^\omega)$, say $U$ spreads onto $V$ if there is some $F:\omega^\omega\rightarrow\omega^\omega$ such that $F(X)\in V$ for every $X\in U$, where $F(X)=\{F(x): x\in X\}$.

The original motivation for this was in the context of working with cardinal characteristics of the continuum without choice - about which I have asked some questions before. This also was a natural, and hopefully less trivial, outgrowth of this question.

However, especially in light of question 2 below, it is unclear whether there's actually any relation. So now it's just a pure curiosity question.

Here's a trivial example. If $A=\{$dominating families$\}$ and $B=\{$escaping families$\}$, then $A$ spreads onto $B$ via the identity map.

Here's a less trivial - maybe even interesting! :P - example. Let $C=\{$pointed perfect sets$\}$ (where a pointed perfect set is a nonempty closed subset $K$ of $\omega^\omega$ with no isolated points, such that every element of $K$ computes the tree representation of $K$ - that is, the set of strings $\sigma\in\omega^{<\omega}$ such that $\exists f\in K(\sigma\prec f)$). Then $C$ spreads onto $A$ (where $A$ is the set of dominating families as above). This argument hinges on two observations:

  • There is a definable way to assign to each perfect $K\subseteq\omega^\omega$ a surjection $s_K: K\rightarrow \omega^\omega$. (Look at the tree representations.)

  • Every real computes only countably many tree representations of perfect sets.

We combine these facts as follows. Given a real $f\in\omega^\omega$, let $T_i$ ($i\in\omega$) be the perfect sets whose tree representation is computable in $f$ (these come ordered by index of Turing reduction), and let $g_i=s_{T_i}(f)$. Then let $F$ be defined as $$F(f)(n)=1+\sum_{i=0}^ns_{T_i}(f)(n).$$ That is, $F$ takes in $f$, and spits out a real dominating each of the countably many reals which $f$ could correspond to in any of the perfect sets $f$ computes.

Now suppose $T$ is any pointed perfect set; I want to argue that $F(T)$ is a dominating family. Since $s_T$ is surjective, it's enough to show that for each $f\in T$, there is some $g\in F(T)$ such that $g$ dominates $s_T(f)$. But since $T$ is perfect, $f$ computes the tree representation of $T$; so we may take $g=F(f)$, and problem solved.

However, each of the two results above has a natural follow-up question:

Question 1. Does $B$ (escaping) spread onto $A$ (dominating)?

Of course the answer is consistently "no," since it is consistent that the dominating number is strictly greater than the bounding number. However, I would like a ZF answer. Less valuable, but still very interesting, would be answers in either ZFC or ZF+AD+etc (which don't decide whether $\mathfrak{b}<\mathfrak{d}$ and which prove a weak CH, respectively - so the question is not trivial in either case).

Question 2. Can we drop "pointed" in the second example? That is, does the family of merely perfect sets spread onto $A$?

I don't see how to do this even in ZFC; we seem to need some way to assign to each real $f$ a family of "few" (specifically, smaller than the escaping number) reals which it needs to dominate. Via pointedness, we were able to cheat - every real is in only countably many pointed perfect sets. However, without this I don't see how to get the argument to go through. (Note that there are perfect sets whose tree representations aren't computed by any of their elements!)

Actually, I'm beginning to think that the right approach for this type of question is to restrict attention to "pointed subsets of $\omega^\omega$" - that is, dominating/escaping/perfect/whatever sets, every element of which computes "the right representation" of the set in question. However, I still don't know quite what I mean by that. Any thoughts on the matter, while not the main focus of this question, would be welcome!


I hope that I didn't misunderstand the question. If escaping family means a family of reals which is unbounded in $\leq^*$ then an Erdős-Sierpinski-type proof gives that in ZFC+CH we have that $B$ spreads onto $A$ as follows:

From an enumeration of the Borel non-dominating sets $\{B'_\alpha:\alpha<\mathfrak{c}\}$ taking unions of the initial segments (and regrouping if necessary) we can get an increasing sequence of non-dominating Borel sets $\{B_\alpha:\alpha<\mathfrak{c}\}$ such that for every $\alpha$ the ring $B_{\alpha+1} \setminus B_\alpha$ has cardinality $\mathfrak{c}$. Do the same with non-escaping Borel sets and obtain the enumeration $\{C'_\alpha:\alpha<\mathfrak{c}\}$ and the increasing sequence $\{C_\alpha:\alpha<\mathfrak{c}\}$ with $|C_{\alpha+1} \setminus C_\alpha|=\mathfrak{c}$. (We can also assume that $|C_0|=|B_0|=\mathfrak{c}$).

Fix bijections $(f_\alpha)_{\alpha<\mathfrak{c}}$ between the sets $B_0$ and $C_0$, and $B_{\alpha+1} \setminus B_\alpha$ and $C_{\alpha+1} \setminus C_\alpha$, respectively. Let $F:\omega^\omega \to \omega^\omega$ be the function which consists of these bijections, i. e., $F(x)=f_\alpha(x)$ iff $x \in B_{\alpha+1} \setminus B_\alpha$.

We claim that a set $S$ is dominating if and only if $F(S)$ is escaping. So suppose that $S$ is dominating but $F(S)$ is not escaping. Then $F(S)$ is covered by a Borel non-escaping set, which must be enumerated in $\{C'_\alpha:\alpha<\mathfrak{c}\}$ therefore $F(S)$ is covered by $C_\alpha$ for some $\alpha<\mathfrak{c}$. But $F^{-1}(C_\alpha)=B_\alpha$ by definition of $F$, so $S \subset B_\alpha$, contradicting the assumption that $S$ was dominating. The other direction is the same.

This proof works in a more general context, namely for $\sigma$-ideals which have cofinality $\mathfrak{c}$.

  • $\begingroup$ Why must $F(S)$ be covered by a Borel nonescaping set? $\endgroup$ Sep 10 '16 at 3:32
  • 1
    $\begingroup$ If a set $X \subset \omega^\omega$ is non-escaping, then it means that there exists a real $f \in \omega^\omega$ so that every element of $X$ is $\leq^* f$, in other words $X \subset \{r: r \leq^*f\}$ where the latter is a Borel set which is non-escaping. $\endgroup$
    – vzoltan
    Sep 10 '16 at 3:37
  • $\begingroup$ Ah, yes, silly me. This is very neat! Of course this isn't a ZFC proof, but still +1! $\endgroup$ Sep 10 '16 at 19:31

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